Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q^2 + 4q}{q^2 + 9q} \times \dfrac{-3q^2 + 3}{q^2 + 5q + 4} $
Explanation: First factor out any common factors. $k = \dfrac{q(q + 4)}{q(q + 9)} \times \dfrac{-3(q^2 - 1)}{q^2 + 5q + 4} $ Then factor the quadratic expressions. $k = \dfrac {q(q + 4)} {q(q + 9)} \times \dfrac {-3(q + 1)(q - 1)} {(q + 1)(q + 4)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {q(q + 4) \times -3(q + 1)(q - 1) } {q(q + 9) \times (q + 1)(q + 4) } $ $k = \dfrac {-3q(q + 1)(q - 1)(q + 4)} {q(q + 1)(q + 4)(q + 9)} $ Notice that $(q + 1)$ and $(q + 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-3q\cancel{(q + 1)}(q - 1)(q + 4)} {q\cancel{(q + 1)}(q + 4)(q + 9)} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $k = \dfrac {-3q\cancel{(q + 1)}(q - 1)\cancel{(q + 4)}} {q\cancel{(q + 1)}\cancel{(q + 4)}(q + 9)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $k = \dfrac {-3q(q - 1)} {q(q + 9)} $ $ k = \dfrac{-3(q - 1)}{q + 9}; q \neq -1; q \neq -4 $